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3 years ago

During which process of the water cycle does water change from a gas to a liquid? A.precipitation

Chemistry

2 answers:

Helga [31]3 years ago

6 0

Answer: Option (B) is the correct answer.

Explanation:

Condensation is defined as a process in which vapor state of a substance changes into liquid state.

Precipitation is defined as a reaction in which two aqueous solutions on chemically combining with each other leads to the formation of an insoluble solid substance.

Evaporation is defined as a process in which liquid state of a substance changes into vapor state.

Transpiration is defined as a process in which plants absorb water with the help of their roots and then release water vapor through the pore present in their leaves.

Therefore, we can conclude that during condensation process of the water cycle, water change from a gas to a liquid.

madreJ [45]3 years ago

3 0

The correct answer is B.Condensation. The other person who answered this question is wrong, it is NOT A. precipitation.

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What is the concentration of a solution in which 10.0 g of AgNO3 is dissolved in 500 mL of solution?

Zigmanuir [339]

molar concentration of AgNO₃ solution = 0.118 mole/L

Explanation:

Because we have the volume of the solution and there is no information about the density of the solution I will asume that you ask for the molar concentration.

number of moles = mass / molecular weight

number of moles of AgNO₃ = 10 / 170 = 0.0588

molar concentration = number of moles / volume (L)

molar concentration of AgNO₃ solution = 0.0588 / 0.5

molar concentration of AgNO₃ solution = 0.118 mole/L

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3 years ago

Which of the following notations is the correct noble gas configuration for Li

Anettt [7]

Noble gas configuration for Li : [He]2s¹

<h3>Further explanation </h3>

In an atom, there are levels of energy in the shell and sub-shell

This energy level is expressed in the form of electron configurations.

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1s²2s¹

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[He]2s¹

8 0

2 years ago

Read 2 more answers

A 125g metal block at a temperature of 93.2 degrees Celsius was immersed in 100g of water at 18.3 degrees Celsius. Given the spe

nikitadnepr [17]

Answer:

$\large \boxed{34.2\, ^{\circ}\text{C}}$

Explanation:

There are two heat transfers involved: the heat lost by the metal block and the heat gained by the water.

According to the Law of Conservation of Energy, energy can neither be destroyed nor created, so the sum of these terms must be zero.

Let the metal be Component 1 and the water be Component 2.

Data:

For the metal:

$m_{1} =\text{125 g; }T_{i} = 93.2 ^{\circ}\text{C; }\\C_{1} = 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

For the water:

$m_{2} =\text{100 g; }T_{i} = 18.3 ^{\circ}\text{C; }\\C_{2} = 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}$

$\begin{array}{rcl}\text{Heat lost by metal + heat gained by water} & = & 0\\q_{1} + q_{2} & = & 0\\m_{1}C_{1}\Delta T_{1} + m_{2}C_{2}\Delta T_{2} & = & 0\\\text{125 g}\times 0.900 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times\Delta T_{1} + \text{100 g} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\Delta \times T_{2} & = & 0\\112.5\Delta T_{1} + 418.4\Delta T_{2} & = & 0\\112.5\Delta T_{1} & = & -418.4\Delta T_{2}\\\Delta T_{1} & = & -3.719\Delta T_{2}\\\end{array}$

$\Delta T_{1} = T_{\text{f}} - 93.2 ^{\circ}\text{C}\\\Delta T_{2} = T_{\text{f}} - 18.3 ^{\circ}\text{C}$

$\begin{array}{rcl}\Delta T_{1} & = & -3.719\Delta T_{2}\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719 (T_{\text{f}} - 18.3 ^{\circ}\text{C})\\T_{\text{f}} - 93.2 ^{\circ}\text{C} & = & -3.719T_{\text{f}} + 68.06 ^{\circ}\text{C}\\4.719T_{\text{f}} & = & 161.3 ^{\circ}\text{C}\\T_{\text{f}} & = & \mathbf{34.2 ^{\circ}}\textbf{C}\\\end{array}\\\text{The final temperature of the block and the water is $\large \boxed{\mathbf{34.2\, ^{\circ}}\textbf{C}}$}$